Question: $ g(x) = \int_{-7}^{x}(t - t^2)\,dt\,$ $ g\,'(10)\, =$
Explanation: The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = t - t^2$ is continuous on $[-7,10]$. Applying the theorem We're given: $ g(x) = \int_{-7}^{x}(t - t^2)\,dt\,$ So the theorem tells us: $ g\,^\prime(x) =x - x^2 $ Evaluating $g'(10)$ $ g'(10)= 10 - 10^2 = -90$ The answer: $g'(2)=-90$